Question

Solubility product of silver bromide $5\times {10}^{-13}$. The quantity of potassium bromide (molar mass taken as $120g{mol}^{-1}$) to be added to $1$ litre of $0.05M$ solution of silver nitrate to start the precipitation of $AgBr$ is:

• A. $6.2\times {10}^{-5}g$
• B. $5.0\times {10}^{-8}g$
• C. $1.2\times {10}^{-10}g$
• D. $1.2\times {10}^{-9}g$

Answer 1

Answer: (D)

$1.2\times {10}^{-9}g$

Let the solubility of $AgBr$ be $S$ $mol{litre}^{-1}$

$AgBr\rightleftharpoons {Ag}^{+}+{Br}^{-}$

Hence, $\left[A{g}^{+}\right]\left[B{r}^{-}\right]=5\times {10}^{-13}$

$\left[A{g}^{+}\right]=0.05$

So $\left[B{r}^{-}\right]={10}^{-11}M$.

It means ${10}^{-11}$ moles of $KBr$ moles in 1 litre of solution.

molecular mass of $KBr=120gmo{l}^{-1}$

So $1.2\times {10}^{-9}g$ of $KBr$ to be added to 1 litre of 0.05M solution of silver nitrate to start precipitation of $AgBr$.